Inverse Laplace Exponent I'm trying to calculate the inverse laplace transform:
$$\large \mathcal{L}^{-1}\left(-\sqrt{\frac ks}e^{-\sqrt{\frac sk}x}\right)$$
I can't figure this out, I tried looking at tables but I couldn't find something that can help me , also I have tried calculating the integral but I didn't manage to get anything..
Hints\help would be appreciated
 A: Well, we are trying to find:
$$\text{y}_\text{k}\left(\text{n}\space;x\right):=\mathscr{L}_\text{s}^{-1}\left[-\sqrt{\frac{\text{k}}{\text{s}}}\cdot\exp\left(-\text{n}\cdot\sqrt{\frac{\text{s}}{\text{k}}}\right)\right]_{\left(x\right)}\tag1$$
Using the linearity of the inverse Laplace transform and the convolution property:
$$\text{y}_\text{k}\left(\text{n}\space;x\right)=\sqrt{\text{k}}\cdot\int_x^0\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\sqrt{\frac{\text{s}}{\text{k}}}\right)\right]_{\left(\sigma\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\sqrt{\text{s}}}\right]_{\left(x-\sigma\right)}\space\text{d}\sigma\tag2$$
It is well known and not hard to prove that:

*

*$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\sqrt{\text{s}}}\right]_{\left(x-\sigma\right)}=\frac{1}{\sqrt{\pi}}\cdot\frac{1}{\sqrt{x-\sigma}}\tag3$$

*$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\sqrt{\frac{\text{s}}{\text{k}}}\right)\right]_{\left(\sigma\right)}=\frac{\text{n}\exp\left(-\frac{\text{n}^2}{4\text{k}\sigma}\right)}{2\sqrt{\text{k}\pi}\sigma^\frac{3}{2}}\tag4$$
So:
$$\text{y}_\text{k}\left(\text{n}\space;x\right)=\frac{\text{n}}{2\pi}\int_x^0\frac{\exp\left(-\frac{\text{n}^2}{4\text{k}\sigma}\right)}{\sigma^\frac{3}{2}}\cdot\frac{1}{\sqrt{x-\sigma}}\space\text{d}\sigma\tag5$$
A: We can also try to dig a bit deeper. Knowing that Laplace Transform of $\operatorname{erf}(-\frac{\sqrt \alpha}{2\sqrt t})$  is $\frac{1}{s}e^{-\sqrt{\alpha s}}$ and taking the first derivative over $\sqrt \alpha$, we may suppose that the desired function has the representation
$$f(t)=\frac{a}{\sqrt t}e^{-\frac{b}{t}}$$
where $a$ and $b$ are some constants.
Performing LT
$$I(s)=\int_0^\infty f(t)e^{-st}dt=\frac{a}{\sqrt s}\int_0^\infty\frac{dx}{\sqrt x}e^{-x-\frac{bs}{x}}=\frac{2a}{\sqrt s}\int_0^\infty e^{-t^2-\frac{bs}{t^2}}dt$$
$$=\frac{2a}{\sqrt s}\int_0^\infty e^{-(t-\frac{\sqrt{bs}}{t})^2}e^{-2\sqrt{bs}}dt$$
Now we can use Glasser's Master Theorem, or just use the substitution $x=\frac{\sqrt{bs}}{t}$ to evaluate the integral:
$$I(s)=\frac{{\sqrt\pi}\,a}{\sqrt s}e^{-2\sqrt{bs}}$$
The last action is to choose the appropriate coefficients $a$ and $b$.
